If the measure of angle $C$ is double the measure of angle $B$, what is the measure of angle $A$ in triangle $ABC$?

[asy]

pair A,B,C;

A=(0,0);
B=(5,0);
C=(-0.8,2.5);

draw(A--B--C--A);

label("$A$",A,SW);
label("$B$",B,SE);
label("$C$",C,N);

draw((4,0.6)..(3.8,0.4)..(3.9,0.1),ArcArrow);

label("$21^\circ$",(4,0.6),E);

[/asy]
Explanation: Since the measure of angle $C$ is twice the measure of angle $B$, $\angle C = 2\cdot 21^\circ = 42^\circ$. It follows that $\angle A = 180^\circ - 21^\circ - 42^\circ = \boxed{117^\circ}$.